step-by-step explanation:
you solve for the inverse of a function algebraically by switching the x and y coordinates and solving for the new y. the first one:
y = 3 + 2 ln(x)
switching x and y:
x = 3 + 2 ln(y) we need to solve for the new y. begin by subtracting 3 from both sides:
x - 3 = 2 ln(y) now divide both sides by 2:
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our goal is to get that y outside the current position in which it's in, so we have to eliminate the ln. do this by raising each side of the equation to e because e and ln "undo" each other:
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since the e and the ln cancel each other out, we are left with:
now for the next one:
![y=\sqrt[3]{(x+4)^2}](/tex.php?f=y=\sqrt[3]{(x+4)^2})
first, let's rewrite that in exponential form cuz it's easier that way!
^{\frac{2}{3}})
now switch x and y:
^{\frac{2}{3}})
we have to get that y out of that set of parenthesis. begin by taking the reciprocal of the power on both sides. the reciprocal of 2/3 is 3/2. multiplying 2/3*3/2 = 1. that's why we do it.
![x^{\frac{3}{2}}=[(y+4)^{\frac{2}{3}}]^{\frac{3}{2}}](/tex.php?f=x^{\frac{3}{2}}=[(y+4)^{\frac{2}{3}}]^{\frac{3}{2}})
the left side is reduced to simply y + 4 now. subtract 4 from both sides:
and in radical form:
the last one is a little tricky because it involves many steps to separate the x and y. i am going to start with the x and y already switched:
begin by multiplying both sides by y + 5. they cancel each other out on the right:
=15y+50)
use the distributive property now:
xy + 5x = 15y + 50
get the y terms together on one side and the other terms on the other side of the equals sign:
5x - 50 = 15y - xy
factor the common y out on the right:
5x - 50 = y(15 - x)
to solve for y, divide both sides by 15 - x:
when it comes to graphing them, you could do one of two things (other than just graphing them on your calculator and copying what you
first, you could simply make an x/y chart for one of the equations, either the function or its inverse. then switch the x and y coordinates to fill in the table for the other. for example, if i plug an x value of 3 into a function (not one of yours above, just some random function) and get a y value of 1, the coordinate for the function is (3, 1). on the inverse it will be (1, 3). that's why when we solve for the inverse we switch the x and the y!
second, you could graph the function using an x/y chart. pick points on the function where the graph goes right through where the grids on the graph paper meet, right at a corner (you don't pick points that fall somewhere in the middle of one of the squares on your graph paper, you pick where the graph of the function goes right through one of the corners of the squares). switch the x and y coordinates and plot the new point and that new point is on the inverse.
regardless of how you choose to graph these, a function and its inverse will always reflect each other through the line y = x. if they don't, you did something wrong.
